Is the answer C.? 22. A potential of 12 V is applied across a parallel-plate capacitor which is constructed of square platesof side 10 cm and separation of 4.0 mm. A dielectric slab of constantK = 2.5 is inserted between the plates so that it completely fills the space. What is the bound surfacecharge density on the dielectric?a.12.3 pC/m²b. 15.9 pC/m²39.8 PC/m²d. 56.8 pC/m²C.e.85.4 pC/m² m222A507) a.4)120-212>E. C = KA² = (2.5) (10x10 m ) (10x10 m ) (8.85 x 10" 1²)E₁₁ = 2EKd22 E = (3000)-(1200) = 1800V = (1800) (4x10 ²³ ) = 7.2 v-10Q = (5.531 x10 "F) (7.2 U) = 3.98 x 10 c39823.2 PC/m²D2 TEOLIn F²d.(4x10-³ m)3Show Markup Toolbar398.232 PC(10 x 10 ²2 ) × ( 10×10 ²³ ) m²-11= 5.531x10 F

Answered: 22. A potential of 12 V is applied…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Is the answer C.?

22. A potential of 12 V is applied across a parallel-plate capacitor which is constructed of square plates
of side 10 cm and separation of 4.0 mm. A dielectric slab of constant
K = 2.5 is inserted between the plates so that it completely fills the space. What is the bound surface
charge density on the dielectric?
a.
12.3 pC/m²
b. 15.9 pC/m²
39.8 PC/m²
d. 56.8 pC/m²
C.
e.
85.4 pC/m²

m2
22
A
50
7) a.
4)
120
-2
12
>
E. C = KA² = (2.5) (10x10 m ) (10x10 m ) (8.85 x 10" 1²)
E₁₁ = 2
EK
d
22 E = (3000)-(1200) = 1800
V = (1800) (4x10 ²³ ) = 7.2 v
-10
Q = (5.531 x10 "F) (7.2 U) = 3.98 x 10 c
39823.2 PC/m²
D
2 TEOL
In F²
d.
(4x10-³ m)
3
Show Markup Toolbar
398.232 PC
(10 x 10 ²2 ) × ( 10×10 ²³ ) m²
-11
= 5.531x10 F

Expert Solution

Step 1

$G i v e n, T h e p o t e n t i a l, V = 12 V T h e s q u a r e h a v i n g s i d e, a = 10 c m = 0.1 m T h e s e p e r a t i o n, d = 4 \times 10^{- 3} m D i e l e c t r i c c o n s \tan t, k = 2.5 σ_{p} = σ_{f} (1 - \frac{1}{k}) S u r f a c e b o u n d f r e e c h a r g e i s, σ_{f} = ε_{0} E E = V / d E = \frac{12}{(4 \times 10^{- 3})} E = 3000 V / m$

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